Square ABCD has sides of length 22 cm. What is the area, in square centimeters, of the region shaded gray? [asy] import olympiad; size(150); pair A = (0, 10), B = (10, 10), E = (10, 7), C = (10, 0), D = (0, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(A--E); draw(D--E); fill(A--B--E--cycle, gray); fill(D--C--E--cycle, gray); label("$A$", A, NW); label("$B$", B, NE); label("$E$", E, NE); label("$C$", C, SE); label("$D$", D, SW); [/asy]
Answer: To find the area of the shaded region, we can instead find the area of the white region, and subtract that area from the total area of the square. The white region is a triangle, so its area is half of its base multiplied by its altitude. We know that the base of $\triangle AED$ is $AD$, and it has length 22, since it is one of the sides of the square. Let us now draw in its altitude: [asy] import olympiad; size(150); pair A = (0, 10), B = (10, 10), E = (10, 7), C = (10, 0), D = (0, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(A--E); draw(D--E); draw((0, 7)--E); fill(A--B--E--cycle, gray); fill(D--C--E--cycle, gray); label("$A$", A, NW); label("$B$", B, NE); label("$E$", E, NE); label("$C$", C, SE); label("$D$", D, SW); [/asy] We can see that this altitude is parallel to $AB$ (because it and $AB$ are both perpendicular to $AD$), so its length must also be equal to that of $AB$ - 22 cm. Thus, the area of $\triangle AED$ is: $\frac{22 \cdot 22}{2} = 11 \cdot 22$. The area of the entire square is $22 \cdot 22$, so the area of the region shaded gray is $22 \cdot 22 - 11 \cdot 22 = 11 \cdot 22 = \boxed{242 \text{ cm}^2}$.